题目:
给你一个单链表的头节点 head ,请你判断该链表是否为回文链表。
如果是,返回 true ;否则,返回 false 。
输入:head = [1,2,2,1]
输出:true
输入:head = [1,2]
输出:false
提示:
链表中节点数目在范围[1, 105] 内
0 <= Node.val <= 9
思考:
快指针走到末尾,慢指针刚好到中间。其中慢指针将前半部分反转。然后比较
pre temp 用于将前半段反转
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public boolean isPalindrome(ListNode head) {
if (head == null || head.next == null) return true;
ListNode slow = head;
ListNode fast = head;
ListNode pre = head;
ListNode temp = null;
while (fast != null && fast.next != null) {
pre = slow;
slow = slow.next;
fast = fast.next.next;
pre.next = temp;
temp = pre;
}
if(fast != null) slow = slow.next;//为了处理链表个数为奇数的情况。
while(pre != null && slow != null) {
if(pre.val != slow.val) {
return false;
}
pre = pre.next;
slow = slow.next;
}
return true;
}
}
LC